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<a class="zhengwen">① Take the bacteria’s photo to get the volume Vold, Wait for the twait and take one photo again to get the volume Vnew</a> | <a class="zhengwen">① Take the bacteria’s photo to get the volume Vold, Wait for the twait and take one photo again to get the volume Vnew</a> | ||
<a class="zhengwen_disblock">② Calculate the growth rate according to the formula.</a><a>$\mu =\dfrac {nV_{nen}-lnV_{old}} {t_{wait}}$</a> | <a class="zhengwen_disblock">② Calculate the growth rate according to the formula.</a><a>$\mu =\dfrac {nV_{nen}-lnV_{old}} {t_{wait}}$</a> | ||
− | <a class="zhengwen_disblock">③ it can be inferred out that the upper limit of the replication folk Nmax is </a><a>$\lceil log_{2}V_{std}e^{\mu(T_{C}+T_{D})} \rceil$<a class="zhengwen_disblock">when the volume of the bac is V</a> | + | <a class="zhengwen_disblock"><br>③ it can be inferred out that the upper limit of the replication folk Nmax is </a><a>$\lceil log_{2}V_{std}e^{\mu(T_{C}+T_{D})} \rceil$<a class="zhengwen_disblock">when the volume of the bac is V</a> |
− | <a class="zhengwen_disblock">④ The volume when the bac complete stage C and getting into stage D:</a><a>$V_{D} =V_{std}e^{\mu T}$</a> | + | <a class="zhengwen_disblock"><br>④ The volume when the bac complete stage C and getting into stage D:</a><a>$V_{D} =V_{std}e^{\mu T}$</a> |
<a class="zhengwen">⑤ the progress of the smallest replication folk</a><a>$$x= | <a class="zhengwen">⑤ the progress of the smallest replication folk</a><a>$$x= | ||
\begin{cases} | \begin{cases} | ||
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$$</a> | $$</a> | ||
<a class="zhengwen_disblock">⑦ Infer how long will the bac split</a><a>$T_{C}+T_{D} - \dfrac {lnV-lnV_{std}}{\mu}$</a> | <a class="zhengwen_disblock">⑦ Infer how long will the bac split</a><a>$T_{C}+T_{D} - \dfrac {lnV-lnV_{std}}{\mu}$</a> | ||
− | <a class="zhengwen_disblock">⑧ Infer how long will it take for the bac to form the haploid chromosome without replication forks if all the Oric is blocked.</a><a>$T_{C}+T_{D}-xT_{C}$</a> | + | <a class="zhengwen_disblock"><br>⑧ Infer how long will it take for the bac to form the haploid chromosome without replication forks if all the Oric is blocked.</a><a>$T_{C}+T_{D}-xT_{C}$</a> |
− | <a class="zhengwen_disblock">⑨ The range of volume cyclical changes:</a><a>$[\dfrac {V_{std}e^{\mu(T_{C}+T_{D})}}{2},V_{std}e^{\mu (T_{C}+T_{D})}]$</a> | + | <a class="zhengwen_disblock"><br>⑨ The range of volume cyclical changes:</a><a>$[\dfrac {V_{std}e^{\mu(T_{C}+T_{D})}}{2},V_{std}e^{\mu (T_{C}+T_{D})}]$</a> |
<a class="zhengwen">⑩ The interval between splits under the current grow environment.</a> | <a class="zhengwen">⑩ The interval between splits under the current grow environment.</a> | ||
<a class="zhengwen">3. when will the normal cells be really effected by Oric</a> | <a class="zhengwen">3. when will the normal cells be really effected by Oric</a> |
Revision as of 10:04, 1 November 2017
③ it can be inferred out that the upper limit of the replication folk Nmax is $\lceil log_{2}V_{std}e^{\mu(T_{C}+T_{D})} \rceil$when the volume of the bac is V
④ The volume when the bac complete stage C and getting into stage D:$V_{D} =V_{std}e^{\mu T}$ ⑤ the progress of the smallest replication folk$$x= \begin{cases} \dfrac {lnV-lnV_{std}} {\mu T_{c}}- \lfloor \dfrac {lnV-lnV_{std}} {ln2} \rfloor \dfrac {ln2} {\mu T_{C}} ,V>V_{std}\\ \\ 0,V < V_{std} \end{cases} $$ ⑥ Replication process of the replication folks$$\begin{cases} \quad \ \ \ x\\ \\ \dfrac {1} {log_{2}e^{\mu T_{c}}} + x\\ \\ \dfrac {2} {log_{2}e^{\mu T_{c}}} + x\\ \\ \quad \ \ \ .\\ \quad \ \ \ .\\ \quad \ \ \ .\\ \\ \dfrac {N_(max)} {log_{2}e^{\mu T_{c}}} + x \end{cases} $$ ⑦ Infer how long will the bac split$T_{C}+T_{D} - \dfrac {lnV-lnV_{std}}{\mu}$
⑧ Infer how long will it take for the bac to form the haploid chromosome without replication forks if all the Oric is blocked.$T_{C}+T_{D}-xT_{C}$
⑨ The range of volume cyclical changes:$[\dfrac {V_{std}e^{\mu(T_{C}+T_{D})}}{2},V_{std}e^{\mu (T_{C}+T_{D})}]$ ⑩ The interval between splits under the current grow environment. 3. when will the normal cells be really effected by Oric Figure 4. When the normal bac is going to form a new replication fork and Oric is blocked at this time, the replication process is really inhibited. so when Oric is unblocked, the time bac is actually effected is the time it takes to form the new replication fork:$T_{r1}=\dfrac {ln2}{\mu}-xT_{C}. $Equal with$T_{r1}=\dfrac {ln2}{\mu}-(\dfrac {lnV-lnV_{std}} {\mu}-\lfloor \dfrac {lnV-lnV_{std}} {ln2} \rfloor \dfrac {ln2} {\mu})$ 4. when will the abnormal cells recover to the normal, predictable stage. According to the Bacteria under the CRISPR control, cell Volume will become many times to normal one, but it only have one chromosome. If we allow its initiate new replication process (release its OriC), it want to initiate many OriC to recovery itself, but previous model didn’t describe this phenom, so we define a new parameter “d” to supply. bac’s rule of growth$V_{new} =V_{old}e^{\mu T}$ bac’s rule of triggering replication innitition$V>2^{N}V_{std}$ reason of division: bac will definitely divide after complete$T_{C}+T_{D}$ The triggercondition to initiate replication under abnormal situation: the shortest interval of forming two replication forks.